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kingwinner
 Post subject: Strictly Convex Functions
PostPosted: Tue, 31 Jan 2012 21:20:33 UTC 
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kingwinner
 Post subject: Re: Strictly Convex Functions
PostPosted: Tue, 31 Jan 2012 21:21:58 UTC 
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So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).
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Shadow
 Post subject: Re: Strictly Convex Functions
PostPosted: Tue, 31 Jan 2012 21:23:27 UTC 
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kingwinner wrote:
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What is your little circle thingy?

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outermeasure
 Post subject: Re: Strictly Convex Functions
PostPosted: Tue, 31 Jan 2012 23:19:51 UTC 
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kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).


Think about the line segment [x,y]. What must f be if you have f(y)-f(x)=(Df)(x)(y-x)?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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kingwinner
 Post subject: Re: Strictly Convex Functions
PostPosted: Wed, 1 Feb 2012 00:54:57 UTC 
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outermeasure wrote:
kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).


Think about the line segment [x,y]. What must f be if you have f(y)-f(x)=(Df)(x)(y-x)?

(The little circle thingy is dot product.)

Let x ≠ y.
Suppose f(y) = f(x) + grad f(x) o (y-x).
By strict convexity, for all 0<s<1,
f[(1-s)x+sy] < (1-s)f(x) + s f(y)
f[x+s(y-x)]< (1-s)f(x) + s [f(x) + grad f(x) o (y-x)]= f(x) + s [grad f(x) o (y-x)]
But where is the contradiction?
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outermeasure
 Post subject: Re: Strictly Convex Functions
PostPosted: Wed, 1 Feb 2012 01:20:30 UTC 
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kingwinner wrote:
outermeasure wrote:
kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).


Think about the line segment [x,y]. What must f be if you have f(y)-f(x)=(Df)(x)(y-x)?

(The little circle thingy is dot product.)

Let x ≠ y.
Suppose f(y) = f(x) + grad f(x) o (y-x).
By strict convexity, for all 0<s<1,
f[(1-s)x+sy] < (1-s)f(x) + s f(y)
f[x+s(y-x)]< (1-s)f(x) + s [f(x) + grad f(x) o (y-x)]= f(x) + s [grad f(x) o (y-x)]
But where is the contradiction?


So you reduce that to univariate case. What is the relationship between secants and tangents there? What does that tell you?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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